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+// Example 22_11
+clc;funcprot(0);
+//Given data
+P=27000;// kW
+p_1=60;// bar
+T_1=450;// °C
+p_v=707.5;// The condenser vaccum in mm of Hg
+p_2=3;//bar
+n_t=87;// The turbine efficiency
+n_b=90;// The boiler efficiency in %
+n_a=95;//The alternator efficiency in %
+n_m=98;//The mechanical efficiency in %
+p_b=760;// cm of Hg
+
+//Calculation
+p_3=((p_b-p_v)/p_b)*1.013;//The condenser pressure bar
+// From h-s chart:
+h_1=3296;// kJ/kg
+h_2a=2606;// kJ/kg
+h_3a=2163;// kJ/kg
+h_2=h_1-((n_t/100)*(h_1-h_2a));// kJ/kg
+h_3=h_2-((n_t/100)*(h_2-h_3a));// kJ/kg
+//From steam tables
+h_f4=162;// kJ/kg (at 0.07 bar)
+h_f5=558;// kJ/kg (at 3 bar)
+//Assume m=y(1)
+function[X]=bled(y)
+    X(1)=((1-y(1))*(h_f5-h_f4))-(y(1)*(h_2-h_f5));
+endfunction
+y=[0.1]
+z=fsolve(y,bled);
+m=z(1);// kg/kg of steam generated
+W=(h_1-h_2)+((1-m)*(h_2-h_3));//Work developed per kg of steam in kJ/kg 
+W_act=(P/((n_a/100)*(n_m/100)));//Actual work developed by the turbine kW
+m_s=(W_act/W)*(3600/1000);// Steam generated per second in tons/hr
+P_p=P*(10/100);// Pump power in kW
+P_net=P*(1-(10/100));// Net power available in kW
+Q_s=((m_s*1000*(h_1-h_f5))/((n_b/100)*3600));// Heat supplied in the boiler in kW
+n_o=(P_net/Q_s)*100;// The overall efficiency of the plant in %
+printf('\n(a)The steam bled per kg of steam supplied to the turbine=%0.3f kg/kg of steam generated \n(b)Steam generated per hour=%0.1f tons/hr \n(c)The overall efficiency of the plant=%0.1f percentage',m,m_s,n_o);
+// The answer vary due to round off error